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Is this the right working out for integral x^3sin(x)dx/[(x^2+1)(x^2+9)]

  1. May 23, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex] int^{infty}_{0} \frac{x^3sinx}{(x^2+1)(x^2+9)} dx [/tex]


    2. Relevant equations

    I understand the value of the integral will be 2 * pi * [sum of residues]
    and I am also aware of the formula

    Res_{z=z_0} f(z) = [tex] \frac{g^{m-1} (z_0)}{(m-1)!} [/tex]


    3. The attempt at a solution

    I know the poles exsit and are x= + / - i and x = +/- 3i

    only 2 in the countour and that is x=i and x=3i

    so I get stuck here because all the examples have a bit of working out with a less than or equal to sign and an absolule value, which shows as it goes to zero R goes to inifity... I just don't understand how to set that bit up and why it is necessary, if you know what i'm talking about, can you please explain it to me?

    Thanks

    Laura
     
  2. jcsd
  3. May 23, 2007 #2

    mjsd

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    Homework Helper

    ok, to cut a long story short...
    you wish to apply Cauchy integral theorem to help you doing this integral...because it seems "hard" to do if we stay on the real line. Therefore the idea is to analytically continue the function onto the complex plane and treat it as a contour integration. Here you have nice theorems like Cauchy integral theorems, Jordan lemma...etc. to help you do things.

    ok, so you are looking for a "closed" contour such that you can apply your theorems. now, however, your initial integral is only from 0 to infty, so you need to make sure that the "extra piece" used to close the contour actually vanishes in the desire limit. (that's where the R->infty bit comes from) let me list the procedure:

    1. goto complex plane x->z
    2. close contour (from a straight line 0->infty, you add an arc of radius R, then return to zero vertically from iR ->0)
    3. wish to make sure the arc bit disappear in the limit of R->infty, the vertical bit may require more work such as change of variables and other tricks (can't remember off top of my head for this specific case)
    4. now to see whether the arc really disappear, you do a "test" on what happen to the integral
    [tex]\int_{\text{arc}}\ldots = \int_{0}^{\pi/2}\ldots d\theta [/tex]
    where [tex]\ldots[/tex] means the integral in polar form (with R and theta instead of z)
    5. you would really want to just look at absolute value because, phases don't matter, only R matters since R->infty eventually
    6. now, if the function/integrand is bounded from above by some function that goes to zero when R becomes large, then you can say this integral involving the "arc" really vanishes (that's where those less than or equal to sign comes from .... i presume)
    7. once, all auxiliary stuffs are proved to be "OK" you can then proceed to solve the problem using those theorems
     
  4. May 23, 2007 #3

    Dick

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    Science Advisor
    Homework Helper

    Split sin(x) up as (exp(ix)-exp(-ix))/(2i) and split the integral into two parts. For the part with exp(ix) in it you can argue the contribution from the extra part of the contour can be ignored if you close it in the upper half plane. For the exp(-ix) part you will want to close in the lower half plane (use the other set of poles and reverse the overall sign since the contour is going in the opposite direction).
     
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